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If the look angles for a pair of satellites are $Az_1, El_1$ and $Az_2, El_2$, what would be the angular separation between them? An associated problem is how do we determine whether and when the two satellites, in slightly different orbits, i.e. slightly different COE (Clasical Orbital Elements), are in radio frequency interference with respect to a given ground antenna. Again a way to determine their angular separation is through the angle of their respective topocentric vectors

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    $\begingroup$ Possible duplicate of astronomy.stackexchange.com/questions/2542/… $\endgroup$
    – user7073
    Jun 21, 2017 at 15:34
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    $\begingroup$ @uhoh Actually, since there are only two points on the sphere (the other point is the center of the sphere), I think this is just a great circle arc, not a spherical triangle. $\endgroup$
    – user7073
    Jun 21, 2017 at 16:23
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    $\begingroup$ @uhoh I was only objecting to your use of the phrase "spherical triangle". In this case, there are only two points on the surface of the sphere, not 3. $\endgroup$
    – user7073
    Jun 22, 2017 at 2:50
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    $\begingroup$ @uhoh Actually, as i noted above, "the other point is the center of the sphere" aka the origin. The other point isn't on the surface of the sphere. In a spherical triangle, all three points are on the surface of the sphere. Not sure it's really worth arguing over though. $\endgroup$
    – user7073
    Jun 22, 2017 at 14:29
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    $\begingroup$ "My interpretation is that he/she just wants the angular separation between the two satellites, as viewed from a single point on Earth. You would not need the altitude to calculate this" This is the correct interpretation. We have a single point on Earth, antenna, (known lat, lon, and height above datum) and we want the angular separation of two satellites, on different orbits, as looked at by the antenna with given look angles. (El, Az). One way to do this would be to compute the topocentric vector from the antenna to each of the satellites and then compute the angle between them. $\endgroup$
    – user169145
    Jun 23, 2017 at 20:09

2 Answers 2

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From "Practical Astronomy with your Calculator or Spreadsheet" by Peter Duffett-Smith, Jonathan Zwart:

Sometimes it is of interest to know what is the angle between two objects in the sky, and this can be calculated very easily provided their equatorial coordinates ($\alpha$,$\delta$) or ecliptic coordinates ($\lambda, \beta$) are known. The formulae are:
cos $d$ = sin $\delta_1$ sin $\delta_2$ + cos $\delta_1$ cos $\delta_2$ cos ($\alpha_1 - \alpha_2$),
or
cos $d$ = sin $\beta_1$ sin $\beta_2$ + cos $\beta_1$ cos $\beta_2$ cos ($\lambda_1 - \lambda_2$),
where $d$ is the angle between the objects whose coordinates are $\alpha_1, \delta_1$ (or $\lambda_1, \beta_1$) and $\alpha_2, \delta_2$ (or $\lambda_2, \beta_2$). Those formulae are exact and mathematically correct for any values of $\alpha, \delta$ or $\lambda, \beta$. However, when $d$ becomes either very small, or close to 180 degrees, your calculator may not have enough precision to return the correct answer, in which case better expressions are
$ d = \sqrt{cos^{2}\delta * \Delta \alpha^{2} + \Delta\delta^2} $
or
$ d = \sqrt{cos^{2}\beta* \Delta \lambda^{2} + \Delta\beta^2} $
where $\Delta\alpha, \Delta\delta$ (or $\Delta\lambda, \Delta\beta$) are the differences in the two coordinates (i.e. $\Delta\alpha = \alpha_1 - \alpha_2$, etc.). These expressions may be used for values of $d$ within about 10 arcminutes of 0 degrees or 180 degrees. Both $\Delta\alpha$ $(\Delta\lambda) $ and $\Delta\delta$ $(\Delta\beta)$ must be expressed in the same units (e.g. arcseconds) and $d$ will then be returned in those units.

The $\beta$ in the $cos^{2}\beta$ above can be either $\beta_1$ or $\beta_2$. It makes little difference, since this formula applies only when d is nearly 0, or nearly 180. Eg, the two points are either very close together, or almost antipodal.

This method should work for RA/Declination, Elev/Azimuth, Lat/Lon, or any similar coordinate system with north and south "poles." (unsure of the general term for these)

I believe this is what you are asking for, as it's generally the most useful in astronomy, attitude determination, etc.

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  • $\begingroup$ This is a really good answer, pointing out the perils of finite precision calculations. $\endgroup$
    – uhoh
    Jun 23, 2017 at 0:30
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    $\begingroup$ @uhoh Thanks, I've actually dealt with that issue in my work, and it's very easy to use the wrong equations. $\endgroup$ Jun 26, 2017 at 13:09
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Indeed, the solution to the problem is found in solving a spherical triangle, as @uhoh states.

Consider this diagram for a general spherical triangle:

enter image description here

In this problem, $C$ is the zenith, and the two points are $A$ and $B$; the co-altitudes of the two points are the arc-lengths $A-C$ and $B-C$. The difference in the two azimuths is the angle $ACB$, and the desired value is the arc-length $A-B$. The sphere portrayed is the celestial sphere; the observer is on the tiny, invisible terrestrial sphere, located at the centre of the celestial sphere, down where the three straight lines $A-A'$, $B-B'$ and $C-C'$ intersect.

So you use the Law of Cosines for spherical triangles...

EDIT as suggested by @Uhoh

If we use lower-case letters to denote the arc-length of the sides of the spherical triangles, and Greek letters to denote the angles at each vertex of the spherical triangle, the Law of Cosines gives: $$\cos(c)=\cos(a)\times \cos(b)+\sin(a) \times \sin(b) \times \cos(\gamma)$$Note: error correctd!

The co-altitude of a point is the altitude measured down from the zenith, rather than up from the horizon. The co-altitude is just 90 degrees minus the altitude..

So:

  1. Find $b$ and $c$ by subtracting the altitudes of each point from $90$ degrees
  2. Find $\gamma$ by subtracting the smaller azimuth from the larger.( If $\gamma$ is bigger than 180 degrees, then use 360-$\gamma$)

  3. Substitute these values into the right-hand -side of the formula above to find $\cos(c)$

  4. Take the inverse cosine (arcos) of this value to find $c$, the angular separation of the two points

EDITED AGAIN

Writing a small Excel spreadsheet to demonstrate an example:

enter image description here

All the angles are in degrees. Excel's trig functions insist on radians, so the RADIANS function changes degrees to radians, and the DEGREES function, naturally, changes radians to degrees.

Here's the actual values for one case:

enter image description here

You can mimic all of these calculations with a scientific calculator (and even skip the degrees/radians conversions.

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  • $\begingroup$ I think for this to be a good SE answer, you need to show the OP explicitly how the law of cosines should be used. With two $(El, Az)$ points, where do those numbers go in the law of cosines, and how to get the answer as an angle rather than another cosine? $\endgroup$
    – uhoh
    Jun 22, 2017 at 4:03
  • $\begingroup$ Downvote: not a spherical triangle-- origin is center of sphere, not on surface of sphere. $\endgroup$
    – user7073
    Jun 22, 2017 at 14:30
  • $\begingroup$ @barrycarter points $A, B, C$ are on the sphere. Angles $\alpha, \beta, \gamma$ are on the sphere. Triangle $ABC$ is a classic spherical triangle. This is really standard stuff. The origin of the sphere is shown for the benefit of us human viewers, who live in three-space. The triangle exists within the two-space of the sphere's surface. Your comment is wrong and misleading. If you have an answer, please post as an answer where you can use mathematics to explain your point, and others can vote on your statements! $\endgroup$
    – uhoh
    Jun 22, 2017 at 15:15
  • $\begingroup$ @uhoh Well, I've provided a link to the correct answer, a link to wikipedia, and offered to discuss it real time. The problem here isn't a mathematical formula: it's understanding what the question is asking. We're talking about the celestial sphere here, not the planetary Earth sphere. It's like finding the angular distance between two stars. If the diagrammed sphere above was the Earth, the satellites would be on its surface. I'll let this sit for a couple days, and hope someone else comes up w/ a better clarification. $\endgroup$
    – user7073
    Jun 22, 2017 at 16:28
  • $\begingroup$ @barrycarter once you have only $(Az, El)$ coordinates, you loose 3 dimensions. There are no physical dimensions or length, only a 2D mathematical surface where distances are measured in degrees. The center of this mathematical sphere has no specific use anymore. Yes, in the real world, satellites are in orbits, but that's not the case for this question. It's a mathematical 2D problem on a sphere, a spherical triangle problem, specifically because the only information you have are angles. Free yourself of the burdens of the real world and embrace non-Euclidian 2D geometry ;) $\endgroup$
    – uhoh
    Jun 22, 2017 at 17:12

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